If C r ≡ C r 25 and C 0 + 5 ∙ C 1 + 9 ∙ C 2 + … + 101 ∙ C 25 = 2 25 ∙ k …

If Cr≡Cr25 and C0+5∙C1+9∙C2+…+101∙C25=225∙k, then k is equal to ____________.

Solution

∑r=0254r+125Cr

=4∑r=025r.25Cr+∑r=025Cr25
=4∑r=125r×25rCr-124+225

=100∑r=125Cr-125+225
=100.224+225

=22550+1

=51.225
So, k=51

Asked in: JEE Main 2020 (09 Jan Shift 2)

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